Prof. Dr. Hamed Hadhoud. Design of Water Tanks: Part (1)

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1 Design of Water Tanks: Part (1) 1

2 Types of Tanks Elevated Tanks Resting on Soil & Underground Tanks

3 Tank Walls Walls Shallow Medium Deep L/ L/ < & L/ >0.5 /L L L L 3

4 Shallow Walls L 1 m L/ (for same continuity conditions, or m for different continuity conditions) LL/ m where m is the continuity factor Loads are transmitted in the vertical direction g w Total Pressure g w Vertical Strip 0.0 orizontal Strip 4

5 L Deep Walls /L (for same continuity conditions, or m / mll for different continuity cond.) where m is the continuity factor /4 Loads are transmitted in the horizontal direction Due to the fixation of the wall to the floor, the lower portion of the wall doesn t deflect horizontally resulting in transmitting part of the load in the vertical direction g w Total Pressure /4 g w Vertical Strip /4 3 g w 4 g w orizontal Strip 5

6 r Cairo University m L m or r m m L L / Grashoff Tables a and b or use equations load factor, dir load factor, L dir P v = a g w (or b g w ) P h = b g w (or a g w ) 4 mll 4 m mll 4 m m 4 m L 4 / Medium Walls L 4 L L/ < & L/ >0.5 (for same continuity conditions, or mll/ m & mll/ m 0. 5 for different continuity cond.) Loads are transmitted in both horizontal and vertical directions Due to the fixation of the wall to the floor, the lower portion of the wall doesn t deflect horizontally resulting in transmitting part of the horizontal load in the vertical direction L /4 P v P h P v P h P h /4 3 4 P h g w Total Pressure Vertical Strip orizontal Strip 6

7 Tank Floors DL = g w t f + floor finishes LL = g w w f = DL + LL r m1 L1 / ml or r m L / m1l1 Grashoff Tables a and b w a = a w f (in the short direction) w b = b w f (in the long direction) g w t f Or use equations load load factor in L factor in L 1 dir dir 4 ml 4 4 m1l 1 ml 4 m1l 1 4 m L m L L 1 L 7

8 Continuity Factors End 1 End Factor (m) Continuous Continuous 0.76 Continuous inged 0.87 orizontal Beam Roof Slab Continuous Free 1.76 Free 8

9 Statical System of Tank Walls orizontal Beam Roof Slab Cantilever 3 m top 4 m top 5 m tie tie tie Column 9

10 Analysis & Design of Elevated Rectangular Tanks: Vertical Sections 1.0 m 1.0 m w w3 w1 OR Top horizontal beam w w3 w1 10

11 Case of Cantilever Wall w w3 w1 Statically Determinate BMD + + NFD + 11

12 Case of Wall Top (Analysis using 3M Equation) R A B A L w1 w3 C w D Twice Statically Indeterminate From Symmetry: M B = M C Applying 3M 0 M B ( L) M B ( L) 6( r 1 r ) w r w1 L r 4 3 w Get M B M B M B w 6 w3 4 6 R A Get R A BMD 1

13 Case of Wall Top (Analysis using Moment Distribution Method) R A A L D Joint B Member BA BC K 0.75 I/L 0.5 I/L B w1 w3 C w D.F. D BA F.E.M. M BA M BC K BA K BA K BC D BC K K BC K BA BC Balance M. -(M BA +M BC )*D BA -(M BA +M BC )*D BC Carry Over M. 0 0 Moment FM1+ Balance M + 0 FM+ Balance M + 0 M B BMD M B w 6 w3 4 6 R A Get R A 13

14 Case of Wall Top (Analysis using Moment Distribution Method) Case K Carry over factor Fixed-Fixed K=I/L 0.5 Fixed-inged K=0.75 I/L 0.0 Symmetrical K=0.5 I/L 0.0 Fixed-end Moments w= g h 3 g h 15 3 g h 0 3 g h g h 14 14

15 Case of Wall Top R A B A L w1 w3 C w D X Maximum Positive BM in the point of zero shear R A w M X m R A X Q 0 R A 1 w X X 1 w X X X 3 Get X Get M m M m Maximum Positive BM in the floor M B M w L 8 1 f M B M f BMD 15

16 Case of Wall Top R A A L D Tension floor B w1 w3 C w T w w 3 4 f R A Maximum Tension wall bottom Get T f T w L 1, wallmax Get T,wall max T,wall max T f Tension wall section with max. +ve BM T m T, wall max X Get T m NFD 16

17 Case of Wall Top Section Type M N A D 1 1 Air-Side Section M m T m Water-Side Section M B T,wall max B 3 4 C 3 Water-Side Section M B T f Critical sections 4 Air-Side Section M f T f 17

18 Analysis & Design of Elevated Rectangular Tanks: orizontal Sections Consider a /4 from wall bottom p1 /4 From symmetry once-statically indeterminate ML T 1 1 M ( L p1l 1 & 1 L T Get M ) M ( L pl ) 6 p1l 4 Design critical sections; &3 Water-side 1&4 Air-side 3 1 pl 4 p 3 T T L p p p1 L1 1 M M 3 4 M B.M.D. M 18 T T 1 1

19 Design of walls as deep beams (in-plane action) The Tank walls will act as beams carrying the tank floor gravity load Those beams are usually deep beams Check the condition for the deep beam 0.8 for simple beam L 0.4 for continuous beam L L L L 19

20 Design of walls as deep beams (in-plane action) y y y T CT CT CT where U A S 0.86 L for simple beams 0.43L for continuous beam@ positive moment 0.37L for continuous beam@ negative moment M y y U CT CT TU f y g s 0.87 in any case A S,min 0

21 Design of walls as deep beams (in-plane action) 1

22 Detailing Vertical Section Water stop construction joint) Deep Beam RFt

23 Detailing Vertical Section Water stop construction joint) Deep Beam RFt 3

24 Detailing Vertical Section Deep Beam RFt 4

25 Detailing Vertical Section Deep Beam RFt 5

26 Detailing Vertical Section Deep Beam RFt 6

27 Detailing Vertical Section Deep Beam RFt 7

28 Detailing orizontal Section 8

29 Detailing orizontal Section 9

30 Example (1) Design and give full details for the conduit shown below 30

31 Loads 31

32 Straining Actions Vertical Strip: Vertical Strip Floor Longitudinal Strip: Floor Longitudinal Strip 3

33 Design of section (1): Design of Critical Sections 33

34 Design of section (): Design of Critical Sections 34 34

35 Design of section (3): Design of Critical Sections 35 35

36 Design of section (4): Design of Critical Sections 1.5*18.5=

37 Design of Critical Sections Design of beam action of wall: Own weight= 5*0.3*= 15 kn/m Floor load=4.5**0.79= 38.7 kn/m Wu= 1.5*( )= 80.6 kn/m M M L U U wu L 80.65,maxve wu L 80.65,maxve Deep beam kN. m kn. m CT m positive negative moment T U ve A A S S,min M y TU f y g s U CT m 1.74 m y m 1.74 m 900 mm y CT CT y 1.74 m y use CT CT m Same for negative moment 0.43L 96.5kN 0.37L 308mm

38 Reinforcement Details

39 Example () Calculate bending moments and normal forces due to the shown water pressures using both 3M equation and moment distribution methods (t f =t w =0.3 m) 5 m 60 kn/m 50kN/m 5 m 39

40 Using 3M equation A D Twice Statically Indeterminate From Symmetry: M B = M C Applying 3M 0 M M M R A B B B (5 5) M B kN. m kn R A (5) 6( ) 5m B 60 kn/m 50kN/m 5m C 40

41 Using moment distribution method Joint Member BA BC B A D K 0.75 I/L 0.5 I/L D.F. F.E.M. D BA M BA K BA K BA K BC D BC M BC K K BC K BA BC kN. m 5m B M BC M 60 kn/m 50kN/m 5m BA 0.0 C Balance M. -( )*0.6 = 5 -( )*0.4= 16,67 Carry Over M. 0 0 Moment M B RA 5 RA 0 kn 6 M BC M BA

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